Ballistic Theory, Progress report...Suitable for 5yo Kids
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From: Paul B. Andersen on 4 Aug 2005 08:25 jgreen(a)seol.net.au skrev: > Paul B. Andersen wrote: > > jgreen(a)seol.net.au skrev: > > > Paul B. Andersen wrote: > > > > jgreen(a)seol.net.au wrote: > > > > > Thanks- I needed a good laugh! > > > > > > > > Fools laugh at what they don't understand. > > > > > > And how big a fool(or coward), does it take to rabbit on about > > > EARTH rotation, when the issue has NOTHING to do with that? > > > We are discussing the rotation of an axle perpendicular to the earth, > > > and the earth rotation has nothing to do with the scenario; > > > ONLY the difference in gravity. > > > > I don't know what YOU are discussing, but what I > > was answering in the posting you responded to was > > this particular challenge defined by Sue: > > | You are of course welcome to advance an opinion > > | about how an axel should behave if it were repeating > > | a geosynchronous clock to the ground or if it were > > | repeating a ground clock to a geosynchronous satellite. > > | Neither you nor Bz seem able to interpret what Einstein's > > | relativity say's the shaft should do. > > > > My answer is what GR "say's the shaft should do". > > And whether you like it or not, the Earth IS rotating. > > And the rotation of the Earth IS relevant to GR's > > prediction of what "the shaft should do". > > > > However, this was not my main point with "one Earth > > rotation" in my scenario, see below. > > > > The apparent paradox is this: > > If the top of the axle (or shaft) rotates at a slower > > rate than the bottom, the axle should be twisted, > > and "wind up" more and more as time passes. > > I show that this isn't so. > > To do that we can compare the number of turns done > > by the bottom and the top of the axle when it points > > in two different directions relative to the distant stars. > > If the number of turns are equal, it will not twist. > > I have - somewhat arbitrarily - chosen to compare the number > > of turns of ends of the axle each time the axle points > > in the same direction, that is after "one Earth rotation". > > > > > I don't expect you to answer-----you cannot! > > > > But I did. Below is my answer again. > > This is what GR say will happen. > > I challenge you to find and point out an inconsistency. > > Your opinion of GR is irrelevant. > > The challenge is to point out an inconsitency > > in GR, showing that there is a real paradox. > > > > Let there be a clock A on the ground at equator. > > Let there be a clock B in geostationary orbit. > > Let both clocks be on the same radius. > > (on the same line through the center of the Erth) > > > > Let A measure the proper duration of one Earth rotation to be T. > > Then, according to GR, B will measure the proper > > duration of one Earth rotation to be longer, T + delta_T. > > > > Let there be an axle between the two clocks. > > Let this axle rotate in such a way that there is no > > mechanical stress in the axle. > > Let the axle rotate N times during one Earth rotation. > > > > A will measure the rotational frequency to be f_g = N/T > > while B will measure it to be f_s = N/(T + delta_T). > > > > So the ground clock will measure the axle to rotate > > faster than the satellite clock will, but both will > > agree that the axle rotates N times per Earth rotation. > > > > frequency * duration = number_of_rotations > > f_g*T = N > > f_s*(T + delta_T) = N > > > > Loosly said: > > "The satellite clock will see the axle rotate slower, > > but for a longer time." > > > > I do not expect you to point out an inconsistency, > > because there are none. > > I do however expect you to laugh at what you don't understands. > > Fools do. Well, Jim? You didn't find any contradictions above, did you? > After all your pathetic attempts to duck the question, it STILL > remains!! Did you ask me a question? What was it? > You claim here that the axle does NOT twist under GR. Since two of the > clocks are ATTACHED to the axle, they MUST read the SAME time. But GR > says the high unconnected (different gravity) one runs differentlty. > This is a CONTRADICTION!!!!!!!!!!!!!!!!!!! of the "theory" Say, what are you talking about? There were but two clocks in my scenario above, and why do you think it would make a difference if they were ATTACHED to the axle? Paul
From: sue jahn on 4 Aug 2005 08:28 "Paul B. Andersen" <paul.b.andersen(a)hia.no> wrote in message news:1123158357.720543.30030(a)g43g2000cwa.googlegroups.com... > > jgreen(a)seol.net.au skrev: > > Paul B. Andersen wrote: > > > jgreen(a)seol.net.au skrev: > > > > Paul B. Andersen wrote: > > > > > jgreen(a)seol.net.au wrote: > > > > > > Thanks- I needed a good laugh! > > > > > > > > > > Fools laugh at what they don't understand. > > > > > > > > And how big a fool(or coward), does it take to rabbit on about > > > > EARTH rotation, when the issue has NOTHING to do with that? > > > > We are discussing the rotation of an axle perpendicular to the earth, > > > > and the earth rotation has nothing to do with the scenario; > > > > ONLY the difference in gravity. > > > > > > I don't know what YOU are discussing, but what I > > > was answering in the posting you responded to was > > > this particular challenge defined by Sue: > > > | You are of course welcome to advance an opinion > > > | about how an axel should behave if it were repeating > > > | a geosynchronous clock to the ground or if it were > > > | repeating a ground clock to a geosynchronous satellite. > > > | Neither you nor Bz seem able to interpret what Einstein's > > > | relativity say's the shaft should do. > > > > > > My answer is what GR "say's the shaft should do". > > > And whether you like it or not, the Earth IS rotating. > > > And the rotation of the Earth IS relevant to GR's > > > prediction of what "the shaft should do". > > > > > > However, this was not my main point with "one Earth > > > rotation" in my scenario, see below. > > > > > > The apparent paradox is this: > > > If the top of the axle (or shaft) rotates at a slower > > > rate than the bottom, the axle should be twisted, > > > and "wind up" more and more as time passes. > > > I show that this isn't so. > > > To do that we can compare the number of turns done > > > by the bottom and the top of the axle when it points > > > in two different directions relative to the distant stars. > > > If the number of turns are equal, it will not twist. > > > I have - somewhat arbitrarily - chosen to compare the number > > > of turns of ends of the axle each time the axle points > > > in the same direction, that is after "one Earth rotation". > > > > > > > I don't expect you to answer-----you cannot! > > > > > > But I did. Below is my answer again. > > > This is what GR say will happen. > > > I challenge you to find and point out an inconsistency. > > > Your opinion of GR is irrelevant. > > > The challenge is to point out an inconsitency > > > in GR, showing that there is a real paradox. > > > > > > Let there be a clock A on the ground at equator. > > > Let there be a clock B in geostationary orbit. > > > Let both clocks be on the same radius. > > > (on the same line through the center of the Erth) > > > > > > Let A measure the proper duration of one Earth rotation to be T. > > > Then, according to GR, B will measure the proper > > > duration of one Earth rotation to be longer, T + delta_T. > > > > > > Let there be an axle between the two clocks. > > > Let this axle rotate in such a way that there is no > > > mechanical stress in the axle. > > > Let the axle rotate N times during one Earth rotation. > > > > > > A will measure the rotational frequency to be f_g = N/T > > > while B will measure it to be f_s = N/(T + delta_T). > > > > > > So the ground clock will measure the axle to rotate > > > faster than the satellite clock will, but both will > > > agree that the axle rotates N times per Earth rotation. > > > > > > frequency * duration = number_of_rotations > > > f_g*T = N > > > f_s*(T + delta_T) = N > > > > > > Loosly said: > > > "The satellite clock will see the axle rotate slower, > > > but for a longer time." > > > > > > I do not expect you to point out an inconsistency, > > > because there are none. > > > I do however expect you to laugh at what you don't understands. > > > Fools do. > > Well, Jim? > You didn't find any contradictions above, did you? Why on earth would Jim waste time trying to explain division by zero to someone that can't make both ends of a shaft turn at the same rate or keep marbles from vanishing in a garden hose? Sue... > > > After all your pathetic attempts to duck the question, it STILL > > remains!! > > Did you ask me a question? > What was it? > > > You claim here that the axle does NOT twist under GR. Since two of the > > clocks are ATTACHED to the axle, they MUST read the SAME time. But GR > > says the high unconnected (different gravity) one runs differentlty. > > This is a CONTRADICTION!!!!!!!!!!!!!!!!!!! of the "theory" > > Say, what are you talking about? > There were but two clocks in my scenario above, > and why do you think it would make a difference > if they were ATTACHED to the axle? > > Paul >
From: Paul B. Andersen on 4 Aug 2005 08:38 sue jahn skrev: > "Paul B. Andersen" <paul.b.andersen(a)hia.no> wrote in message news:1123097611.454902.209960(a)z14g2000cwz.googlegroups.com... > > > > Sue... skrev: > > > Paul B. Andersen wrote: > > > > sue jahn skrev: > > > > > "Paul B. Andersen" <paul.b.andersen(a)deletethishia.no> wrote in message news:dclsna$41n$1(a)dolly.uninett.no... > > > > > > sue jahn wrote: > > > > > > > You are of course welcome to advance an opinion > > > > > > > about how an axel should behave if it were repeating > > > > > > > a geosynchronous clock to the ground or if it were > > > > > > > repeating a ground clock to a geosynchronous satellite. > > > > > > > Neither you nor Bz seem able to interpret what Einstein's > > > > > > > relativity say's the shaft should do. > > > > > > > > > > > > Why do you think I should have any problem with this? > > > > > > This is yet another old non paradox. > > > > > > > > > > > > Let there be a clock A on the ground at equator. > > > > > > Let there be a clock B in geostationary orbit. > > > > > > Let both clocks be on the same radius. > > > > > > > > > > > > Let A measure the proper duration of one Earth rotation to be T. > > > > > > Then, as you now know and have accepted is experimentally > > > > > > verified for clocks in GPS orbit, B will measure the proper > > > > > > duration of one Earth rotation to be longer, T + delta_T. > > > > > > > > > > > > Let there be an axle between the two clocks. > > > > > > Let this axle rotate in such a way that there is no > > > > > > mechanical stress in the axle. > > > > > > Let the axle rotate N times during one Earth rotation. > > > > > > > > > > > > A will measure the rotational frequency to be f_g = N/T > > > > > > while B will measure it to be f_s = N/(T + delta_T). > > > > > > > > > > > > So the ground clock will measure the axle to rotate > > > > > > faster than the satellite clock will, but both will > > > > > > agree that the axle rotates N times per Earth rotation. > > > > > > > > > > > > frequency * duration = number_of_rotations > > > > > > f_g*T = N > > > > > > f_s*(T + delta_T) = N > > > > > > > > > > > > Loosly said: > > > > > > "The satellite clock will see the axle rotate slower, > > > > > > but for a longer time." > > > > > > > > Sue, I have responded to your challenge and explained > > > > "what Einstein's relativity say's the shaft should do." > > > > > > > > Now I challenge you to point out an inconsitency in the above. > > > > Well, Sue? > > You are not up to it, are you? Thought so. > I can do it with a bag of marbles and two garden hoses and > nobody questions the result. > You broke your proof. > You can fix your proof. And this is the best you can do? :-) It wasn't even witty. So we can conclude that you are unable to point out any inconsistencies in "what Einstein's relativity says the shaft should do." Paul
From: Paul B. Andersen on 4 Aug 2005 09:13 sue jahn skrev: > "Paul B. Andersen" <paul.b.andersen(a)hia.no> wrote in message news:1123158357.720543.30030(a)g43g2000cwa.googlegroups.com... > > > > jgreen(a)seol.net.au skrev: > > > Paul B. Andersen wrote: > > > > jgreen(a)seol.net.au skrev: > > > > > Paul B. Andersen wrote: > > > > > > jgreen(a)seol.net.au wrote: > > > > > > > Thanks- I needed a good laugh! > > > > > > > > > > > > Fools laugh at what they don't understand. > > > > > > > > > > And how big a fool(or coward), does it take to rabbit on about > > > > > EARTH rotation, when the issue has NOTHING to do with that? > > > > > We are discussing the rotation of an axle perpendicular to the earth, > > > > > and the earth rotation has nothing to do with the scenario; > > > > > ONLY the difference in gravity. > > > > > > > > I don't know what YOU are discussing, but what I > > > > was answering in the posting you responded to was > > > > this particular challenge defined by Sue: > > > > | You are of course welcome to advance an opinion > > > > | about how an axel should behave if it were repeating > > > > | a geosynchronous clock to the ground or if it were > > > > | repeating a ground clock to a geosynchronous satellite. > > > > | Neither you nor Bz seem able to interpret what Einstein's > > > > | relativity say's the shaft should do. > > > > > > > > My answer is what GR "say's the shaft should do". > > > > And whether you like it or not, the Earth IS rotating. > > > > And the rotation of the Earth IS relevant to GR's > > > > prediction of what "the shaft should do". > > > > > > > > However, this was not my main point with "one Earth > > > > rotation" in my scenario, see below. > > > > > > > > The apparent paradox is this: > > > > If the top of the axle (or shaft) rotates at a slower > > > > rate than the bottom, the axle should be twisted, > > > > and "wind up" more and more as time passes. > > > > I show that this isn't so. > > > > To do that we can compare the number of turns done > > > > by the bottom and the top of the axle when it points > > > > in two different directions relative to the distant stars. > > > > If the number of turns are equal, it will not twist. > > > > I have - somewhat arbitrarily - chosen to compare the number > > > > of turns of ends of the axle each time the axle points > > > > in the same direction, that is after "one Earth rotation". > > > > > > > > > I don't expect you to answer-----you cannot! > > > > > > > > But I did. Below is my answer again. > > > > This is what GR say will happen. > > > > I challenge you to find and point out an inconsistency. > > > > Your opinion of GR is irrelevant. > > > > The challenge is to point out an inconsitency > > > > in GR, showing that there is a real paradox. > > > > > > > > Let there be a clock A on the ground at equator. > > > > Let there be a clock B in geostationary orbit. > > > > Let both clocks be on the same radius. > > > > (on the same line through the center of the Erth) > > > > > > > > Let A measure the proper duration of one Earth rotation to be T. > > > > Then, according to GR, B will measure the proper > > > > duration of one Earth rotation to be longer, T + delta_T. > > > > > > > > Let there be an axle between the two clocks. > > > > Let this axle rotate in such a way that there is no > > > > mechanical stress in the axle. > > > > Let the axle rotate N times during one Earth rotation. > > > > > > > > A will measure the rotational frequency to be f_g = N/T > > > > while B will measure it to be f_s = N/(T + delta_T). > > > > > > > > So the ground clock will measure the axle to rotate > > > > faster than the satellite clock will, but both will > > > > agree that the axle rotates N times per Earth rotation. > > > > > > > > frequency * duration = number_of_rotations > > > > f_g*T = N > > > > f_s*(T + delta_T) = N > > > > > > > > Loosly said: > > > > "The satellite clock will see the axle rotate slower, > > > > but for a longer time." > > > > > > > > I do not expect you to point out an inconsistency, > > > > because there are none. > > > > I do however expect you to laugh at what you don't understands. > > > > Fools do. > > > > Well, Jim? > > You didn't find any contradictions above, did you? I note with interest that Sue still don't know that the Earth is rotating, and thus believe that division by "one Earth rotation" is division by zero: > Why on earth would Jim waste time trying to explain division by zero > to someone that can't make both ends of a shaft turn at > the same rate or keep marbles from vanishing in a garden > hose? > > Sue... Why, indeed. Since it was you who challenged me to show "what Einstein's relativity says the shaft should do" one would expect that it was you rather than Jim that would point out the inconsistencies you claim are there. You have now throughly demonstrated your inability to do so. Paul
From: bz on 4 Aug 2005 09:39
jgreen(a)seol.net.au wrote in news:1123155872.920713.291420 @g43g2000cwa.googlegroups.com: > No one has explained when I asked here why circular logic isn't > involved here to- temperatures of suns are derived from their > spectrums, right? I hope I did that in my last article. We tell the red/blue shift from the spectrum lines and correct for it. > But last time I looked, elements Black body radiation is independent of the material making up the black body. It can be any element(s). > can give off very > different wavelengths according to their temperatures......circular > again.......... > a neat java program to plot distributions from temperature. http://csep10.phys.utk.edu/guidry/java/planck/planck.html -- bz please pardon my infinite ignorance, the set-of-things-I-do-not-know is an infinite set. bz+sp(a)ch100-5.chem.lsu.edu remove ch100-5 to avoid spam trap |